結果
| 問題 |
No.1357 Nada junior high school entrance examination 3rd day
|
| コンテスト | |
| ユーザー |
 qwewe
|
| 提出日時 | 2025-05-14 13:14:16 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 1,043 ms / 2,000 ms |
| コード長 | 7,897 bytes |
| コンパイル時間 | 307 ms |
| コンパイル使用メモリ | 82,288 KB |
| 実行使用メモリ | 105,680 KB |
| 最終ジャッジ日時 | 2025-05-14 13:15:44 |
| 合計ジャッジ時間 | 15,942 ms |
|
ジャッジサーバーID （参考情報） |
judge5 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 21 |
ソースコード
import sys
# Set a higher recursion depth, just in case. With iterative NTT, it might not be strictly necessary.
# sys.setrecursionlimit(200000)
def solve():
K = int(sys.stdin.readline())
MOD = 998244353
# The problem asks for coefficients c_0, c_1, ..., c_{2K}.
# The maximum index is 2K. The total number of coefficients is 2K+1.
MAX_N = 2 * K
# Precomputation requires factorials up to 2K+1.
# Because the polynomial A(t) has coefficients a_n = 1/(n+1)!,
# for the highest degree term needed n=2K, we need a_{2K} = 1/(2K+1)!.
# So we need factorials and inverse factorials up to 2K+1.
MAX_FACT = MAX_N + 1
# Handle K=0 case if it were possible. The problem states K >= 1.
# If K=0, MAX_N = 0, MAX_FACT = 1. The sum is empty, result is 0. Output should be just c_0=0.
if K == 0:
print("0")
return
# Precompute factorials and their modular inverses.
# Size of arrays needs to be MAX_FACT + 1 to store indices 0..MAX_FACT.
fact = [1] * (MAX_FACT + 1)
inv_fact = [1] * (MAX_FACT + 1)
for i in range(1, MAX_FACT + 1):
fact[i] = (fact[i-1] * i) % MOD
# Compute modular inverse using Fermat's Little Theorem for the largest factorial needed.
# (MOD - 2) is the exponent for modular inverse because MOD is prime.
inv_fact[MAX_FACT] = pow(fact[MAX_FACT], MOD - 2, MOD)
# Compute other inverse factorials iteratively using inv_fact[i] = inv_fact[i+1] * (i+1)
for i in range(MAX_FACT - 1, -1, -1):
inv_fact[i] = (inv_fact[i+1] * (i+1)) % MOD
# Iterative Number Theoretic Transform (NTT)
# This implementation uses Cooley-Tukey Radix-2 DIT FFT algorithm adapted for modular arithmetic.
def ntt(a, inv):
n = len(a)
if n == 1:
return
# Bit-reversal permutation: arrange elements in order required for iterative FFT
j = 0
for i in range(1, n):
rev = n >> 1
while j >= rev:
j -= rev
rev >>= 1
j += rev
if i < j:
a[i], a[j] = a[j], a[i]
# Main FFT loop
len_ = 2 # Current length of DFT sequences being combined
while len_ <= n:
# Root of unity: g=3 is primitive root for MOD 998244353
# w_len is omega_len_ = g^((MOD-1)/len_) mod MOD
w_len = pow(3, (MOD - 1) // len_, MOD)
if inv:
# For inverse NTT, use inverse root of unity
w_len = pow(w_len, MOD - 2, MOD)
half_len = len_ >> 1
for i in range(0, n, len_): # Iterate through blocks of size len_
w = 1 # Current power of root of unity: omega_len_^j
for j in range(half_len): # Butterfly operations
u = a[i + j] # Even indexed element
v = (w * a[i + j + half_len]) % MOD # Odd indexed element multiplied by root power
a[i + j] = (u + v) % MOD
a[i + j + half_len] = (u - v + MOD) % MOD # Ensure result is non-negative
w = (w * w_len) % MOD # Update root power for next step
len_ <<= 1 # Double the length for next stage
if inv:
# Scale by 1/n for inverse transform
n_inv = pow(n, MOD - 2, MOD)
for i in range(n):
a[i] = (a[i] * n_inv) % MOD
# Polynomial inverse computation using NTT (based on Newton-Raphson iteration)
def poly_inv(A, deg):
# Computes polynomial B such that A * B = 1 (mod x^deg)
# Pad A to ensure it has length at least deg
A_padded = A[:deg] + [0] * max(0, deg - len(A))
B = [0] * deg # Result polynomial B
if deg == 0: return B
# Base case: B = 1/A[0] (mod x^1)
B[0] = pow(A_padded[0], MOD - 2, MOD)
# Iteratively double the precision `k`
k = 1
while k < deg:
k *= 2
# Determine required NTT length: smallest power of 2 >= 2k
# This is needed because multiplication of two polynomials of degree k-1 results in degree 2k-2.
ntt_len = 1 << (2*k -1).bit_length() if k > 0 else 1
# Prepare A_k and B_k for NTT: Take required parts and pad to ntt_len
# We only need A up to degree k-1 for computation mod x^k
A_k = A_padded[:min(k, deg)] + [0] * (ntt_len - min(k, deg))
# Use current approximation B up to degree k-1
B_k = B[:min(k, deg)] + [0] * (ntt_len - min(k, deg))
ntt(A_k, False)
ntt(B_k, False)
# Compute C_ntt = NTT(A * B^2) efficiently in frequency domain
# C_ntt[i] = A_k[i] * B_k[i] * B_k[i]
C_ntt = [0] * ntt_len
for i in range(ntt_len):
C_ntt[i] = (A_k[i] * B_k[i]) % MOD
C_ntt[i] = (C_ntt[i] * B_k[i]) % MOD
ntt(C_ntt, True) # Inverse NTT to get coefficients of A*B^2
# Update B using the Newton-Raphson formula: B_new = 2B - A*B^2 (mod x^k)
# We need to update B up to degree k-1. Copy the relevant part of B first.
B_new_k = B[:k]
for i in range(min(k, deg)): # Compute updates up to degree min(k, deg) - 1
term_2B = (2 * B[i]) % MOD
# C_ntt holds coefficients of A*B^2. Indices correspond to powers of x.
B_new_k[i] = (term_2B - C_ntt[i] + MOD) % MOD
# Update B for next iteration. Store the result padded to degree `deg`.
B = B_new_k + [0] * max(0, deg - k)
return B[:deg] # Return the inverse polynomial up to degree deg-1
# Step 1: Define polynomial A(t) coefficients a_n = 1 / (n+1)!
# We need polynomial of degree 2K, so length is 2K+1.
N_poly = MAX_N + 1
A = [0] * N_poly
for n in range(N_poly):
# A[n] is the coefficient of t^n
A[n] = inv_fact[n+1] # Access precomputed inverse factorials: 1/(n+1)!
# Step 2: Compute B(t) = A(t)^(-1) mod t^(N_poly)
# The coefficients b_n of B(t) are B_n / n! (where B_n are Bernoulli numbers with B1=-1/2 convention)
inv_deg = N_poly # Compute inverse modulo t^(2K+1)
B_coeffs = poly_inv(A, inv_deg)
# Step 3: Compute the final coefficients c_i based on the derived formula
# c_{2a} = (2a-1) * (-1)^(a-1) * (B_{2a} / (2a)!) * 2^(2a-1)
C = [0] * (MAX_N + 1) # Result array C[0]...C[2K]
# Precompute powers of 2 modulo MOD
pow2 = [1] * (MAX_N + 1)
for i in range(1, MAX_N + 1):
pow2[i] = (pow2[i-1] * 2) % MOD
# Calculate coefficients c_{2a} for a = 1 to K
for a in range(1, K + 1):
idx = 2*a # Index in C array corresponds to power of pi (pi^idx)
# Safety check: Index should be within computed range
if idx >= N_poly: continue
term1 = (2*a - 1) # The factor (2a-1)
term2 = 1 # The factor (-1)^(a-1)
if (a - 1) % 2 == 1: # Check if a-1 is odd (i.e., a is even)
term2 = MOD - 1 # Represents -1 mod P
# b_{2a} = B_{2a} / (2a)! is the coefficient from poly inverse B_coeffs[idx]
b_2a = B_coeffs[idx]
idx_pow2 = idx - 1 # Index for power of 2 is 2a-1
# Safety check: index for power of 2 must be non-negative
if idx_pow2 < 0 : continue
term4 = pow2[idx_pow2] # Factor 2^(2a-1)
# Combine all terms modulo MOD
c_2a = (term1 % MOD * term2) % MOD
c_2a = (c_2a * b_2a) % MOD
c_2a = (c_2a * term4) % MOD
C[idx] = c_2a # Store the computed coefficient c_{2a} at index 2a
# Step 4: Print the result coefficients C[0], C[1], ..., C[2K] separated by spaces
print(*(C))
solve()